我一直在阅读c代码here。我不明白这种typedef：
typedef const char * (*Responder)( int p1);
p1是响应者的参数吗？ Responder是const char吗？
答案 0 :(得分：6)
typedef（语法上）除了任何其他存储类说明符（例如static或extern）之外别无其他，只要想想没有typedef的情况下会声明的内容：
const char * (*Responder)( int p1);
Responder是指向函数的指针（获取int）返回指向const char的指针。
所以，使用typedef，你有：
Responder是指向函数的指针的类型同义词（获取int）返回指向const char的指针。
相关：http://c-faq.com/decl/spiral.anderson.html
HTH
答案 1 :(得分：2)
使用带有函数指针的typedef绝对可行，如here和here所示。
例如（来自其中一个链接）：
int do_math(float arg1, int arg2) {
return arg2;
}
int call_a_func(int (*call_this)(float, int)) {
int output = call_this(5.5, 7);
return output;
}
int final_result = call_a_func(&do_math);
此代码可以使用typedef重写，如下所示：
typedef int (*MathFunc)(float, int);
int do_math(float arg1, int arg2) {
return arg2;
}
int call_a_func(MathFunc call_this) {
int output = call_this(5.5, 7);
return output;
}
int final_result = call_a_func(&do_math);
在您的特定情况下，它是指向名为Responder的函数的指针，该函数返回char *并将int作为参数。
答案 2 :(得分：1)
typedef只是另一种数据类型的类型别名。几年前我问了一个类似的问题，并指出了以下（很棒的）教程，它将帮助你理解任何复杂的typedef。如果网站出现故障，我会将其添加到参考中。
使用左右规则，“responder是一个指向函数的指针的别名，该函数对参数采用int，并返回指向const char”的指针。
示例列表
The "right-left" rule is a completely regular rule for deciphering C
declarations. It can also be useful in creating them.
First, symbols. Read
* as "pointer to" - always on the left side
[] as "array of" - always on the right side
() as "function returning" - always on the right side
as you encounter them in the declaration.
STEP 1
------
Find the identifier. This is your starting point. Then say to yourself,
"identifier is." You've started your declaration.
STEP 2
------
Look at the symbols on the right of the identifier. If, say, you find "()"
there, then you know that this is the declaration for a function. So you
would then have "identifier is function returning". Or if you found a
"[]" there, you would say "identifier is array of". Continue right until
you run out of symbols *OR* hit a *right* parenthesis ")". (If you hit a
left parenthesis, that's the beginning of a () symbol, even if there
is stuff in between the parentheses. More on that below.)
STEP 3
------
Look at the symbols to the left of the identifier. If it is not one of our
symbols above (say, something like "int"), just say it. Otherwise, translate
it into English using that table above. Keep going left until you run out of
symbols *OR* hit a *left* parenthesis "(".
Now repeat steps 2 and 3 until you've formed your declaration. Here are some
examples:
int *p[];
1) Find identifier. int *p[];
^
"p is"
2) Move right until out of symbols or right parenthesis hit.
int *p[];
^^
"p is array of"
3) Can't move right anymore (out of symbols), so move left and find:
int *p[];
^
"p is array of pointer to"
4) Keep going left and find:
int *p[];
^^^
"p is array of pointer to int".
(or "p is an array where each element is of type pointer to int")
Another example:
int *(*func())();
1) Find the identifier. int *(*func())();
^^^^
"func is"
2) Move right. int *(*func())();
^^
"func is function returning"
3) Can't move right anymore because of the right parenthesis, so move left.
int *(*func())();
^
"func is function returning pointer to"
4) Can't move left anymore because of the left parenthesis, so keep going
right. int *(*func())();
^^
"func is function returning pointer to function returning"
5) Can't move right anymore because we're out of symbols, so go left.
int *(*func())();
^
"func is function returning pointer to function returning pointer to"
6) And finally, keep going left, because there's nothing left on the right.
int *(*func())();
^^^
"func is function returning pointer to function returning pointer to int".
As you can see, this rule can be quite useful. You can also use it to
sanity check yourself while you are creating declarations, and to give
you a hint about where to put the next symbol and whether parentheses
are required.
Some declarations look much more complicated than they are due to array
sizes and argument lists in prototype form. If you see "[3]", that's
read as "array (size 3) of...". If you see "(char *,int)" that's read
as "function expecting (char *,int) and returning...". Here's a fun
one:
int (*(*fun_one)(char *,double))[9][20];
I won't go through each of the steps to decipher this one.
Ok. It's:
"fun_one is pointer to function expecting (char *,double) and
returning pointer to array (size 9) of array (size 20) of int."
As you can see, it's not as complicated if you get rid of the array sizes
and argument lists:
int (*(*fun_one)())[][];
You can decipher it that way, and then put in the array sizes and argument
lists later.
Some final words:
It is quite possible to make illegal declarations using this rule,
so some knowledge of what's legal in C is necessary. For instance,
if the above had been:
int *((*fun_one)())[][];
it would have been "fun_one is pointer to function returning array of array of
^^^^^^^^^^^^^^^^^^^^^^^^
pointer to int". Since a function cannot return an array, but only a
pointer to an array, that declaration is illegal.
Illegal combinations include:
[]() - cannot have an array of functions
()() - cannot have a function that returns a function
()[] - cannot have a function that returns an array
In all the above cases, you would need a set of parens to bind a *
symbol on the left between these () and [] right-side symbols in order
for the declaration to be legal.
Here are some legal and illegal examples:
int i; an int
int *p; an int pointer (ptr to an int)
int a[]; an array of ints
int f(); a function returning an int
int **pp; a pointer to an int pointer (ptr to a ptr to an int)
int (*pa)[]; a pointer to an array of ints
int (*pf)(); a pointer to a function returning an int
int *ap[]; an array of int pointers (array of ptrs to ints)
int aa[][]; an array of arrays of ints
int af[](); an array of functions returning an int (ILLEGAL)
int *fp(); a function returning an int pointer
int fa()[]; a function returning an array of ints (ILLEGAL)
int ff()(); a function returning a function returning an int
(ILLEGAL)
int ***ppp; a pointer to a pointer to an int pointer
int (**ppa)[]; a pointer to a pointer to an array of ints
int (**ppf)(); a pointer to a pointer to a function returning an int
int *(*pap)[]; a pointer to an array of int pointers
int (*paa)[][]; a pointer to an array of arrays of ints
int (*paf)[](); a pointer to a an array of functions returning an int
(ILLEGAL)
int *(*pfp)(); a pointer to a function returning an int pointer
int (*pfa)()[]; a pointer to a function returning an array of ints
(ILLEGAL)
int (*pff)()(); a pointer to a function returning a function
returning an int (ILLEGAL)
int **app[]; an array of pointers to int pointers
int (*apa[])[]; an array of pointers to arrays of ints
int (*apf[])(); an array of pointers to functions returning an int
int *aap[][]; an array of arrays of int pointers
int aaa[][][]; an array of arrays of arrays of ints
int aaf[][](); an array of arrays of functions returning an int
(ILLEGAL)
int *afp[](); an array of functions returning int pointers (ILLEGAL)
int afa[]()[]; an array of functions returning an array of ints
(ILLEGAL)
int aff[]()(); an array of functions returning functions
returning an int (ILLEGAL)
int **fpp(); a function returning a pointer to an int pointer
int (*fpa())[]; a function returning a pointer to an array of ints
int (*fpf())(); a function returning a pointer to a function
returning an int
int *fap()[]; a function returning an array of int pointers (ILLEGAL)
int faa()[][]; a function returning an array of arrays of ints
(ILLEGAL)
int faf()[](); a function returning an array of functions
returning an int (ILLEGAL)
int *ffp()(); a function returning a function
returning an int pointer (ILLEGAL)
<强>参考
<http://ieng9.ucsd.edu/~cs30x/rt_lt.rule.html>